Question

A series combination of two resistors, 7.25 ω and 4.03 ω, is connected to a 9.00 v battery.

Answer 1

a. $11.28\Omega$

The equivalent resistance of a series combination of two resistors is equal to the sum of the individual resistances:

$R_{eq}=R_1 + R_2$

In this circuit, we have

$R_1 = 7.25 \Omega\\R_2 = 4.03 \Omega$

Therefore, the equivalent resistance is

$R_{eq}=7.25 \Omega + 4.03 \Omega=11.28 \Omega$

b. 5.8 V, 3.2 V

First of all, we need to determine the current flowing through each resistor, which is given by Ohm's law:

$I=\frac{V}{R_{eq}}$

where V = 9.00 V and $R_{eq}=11.28 \Omega$. Substituting,

$I=\frac{9.00 V}{11.28 \Omega}=0.8 A$

Now we can calculate the potential difference across each resistor by using Ohm's law again:

$V_1 = I R_1 = (0.8 A)(7.25 \Omega)=5.8 V$

$V_2 = I R_2 = (0.8 A)(4.03 \Omega)=3.2 V$